leetcode 629. K Inverse Pairs Array

629. K Inverse Pairs Array

• User Accepted: 81
• User Tried: 363
• Total Accepted: 83
• Total Submissions: 1074
• Difficulty: Medium

Given two integers n and k, find how many different arrays consist of numbers from 1 to n such that there are exactly k inverse pairs.

We define an inverse pair as following:For ith and jth element in the array, if i < j and a[i] > a[j] then it's an inverse pair; Otherwise, it's not.

Since the answer may very large, the answer should be modulo 10⁹ + 7.

Example 1:

Input: n = 3, k = 0
Output: 1
Explanation: 
Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pair.

Example 2:

Input: n = 3, k = 1
Output: 2
Explanation: 
The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.

Note:

1. The integer n is in the range [1, 1000] and k is in the range [0, 1000].

思路不难想，dp[i][j]表示前i个数字构成j个逆序对的方法数。最朴素的想法是三重循环做，但是可以优化掉最内层对i-1的那层循环。

问题是对边界的处理真的烦，之前的codeM也是，很多题的思路都是对的，但就是边界的问题或者是写法不好导致题目过不去，总是差那么一点。。。真的伤心。。。

public class Solution {  
    public int kInversePairs(int n, int k) {  
        long[][] dp = new long[n+1][k+1];  
          
        for(int i=0; i<=n; i++)  
            dp[i][0] = 1;  
          
        for(int i=1; i<=n; i++)  
            for(int j=1; j<=k; j++) {  
                dp[i][j] = dp[i][j-1]+dp[i-1][j];  
                if(j-i >= 0) dp[i][j] -= dp[i-1][j-i];  
                dp[i][j] += 1000000007;  
                dp[i][j] %= 1000000007;  
            }  
          
        return (int) dp[n][k];  
    }  
}